pH of a saturated solution of  Ca(OH)_2  is 9. The solubility product (K_{sp}) of Ca(OH)_2 is:

  • Option 1)

    0.5\times 10^{-15}

  • Option 2)

    0.25\times 10^{-10}

  • Option 3)

    0.125\times 10^{-15}

  • Option 4)

    0.5\times 10^{-10}

Answers (1)

Option 1

Ca(OH)_2 \leftrightharpoons Ca^{2+} + 2OH^-

Given pH = 9

So, 

pH + pOH = 14

\Rightarrow pOH = 14 -9 =5

\Rightarrow -\log[OH^-] =5

\Rightarrow [OH^-] = 10^{-5}

So,

[Ca^{2+}] = \frac{[OH^-]}{2} = \frac{10^{-5}}{2}

And K_{sp} will be

    K_{sp} = [Ca^{2+}][OH^-]^2

    K_{sp} = \frac{10^{-5}}{2}\times (10^{-5})^2

    K_{sp} = 0.5\times 10^{-15} mol^3/L^3


Option 1)

0.5\times 10^{-15}

Option 2)

0.25\times 10^{-10}

Option 3)

0.125\times 10^{-15}

Option 4)

0.5\times 10^{-10}

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